I was reading Rudin's Real and Complex Analysis. I can understand that if $\phi(E)\triangleq \int_Ef\,d\mu$ then $\phi$ is a measure. However, I'm not so sure how to prove $\int_X g\,d\phi=\int_X gf\,d\mu.$ ($f,g:X\to[0, \infty]$ are measurable functions, and $\mu$ is a measure.) I list my attempted proof below, and would appreciate it if someone can confirm if it's correct, or point out where I'm mistaken, or give a simpler proof.
My proof: Since $g$ is measurable, there exists a sequence $\{s_n\}$ of increasing simple measurable functions on $X$ that converge to $g$, i.e. $0\le s_1\le s_2 \le \cdots$ and $s_n \to g.$ So, by Monotone Convergence Theorem (MCT), $\int_X s_n\,d\phi \to \int_X g\,d\phi.$
Suppose $s_n=\sum_i\alpha_i\chi_{E_i},$ where $\chi_{E_i}$ is the characteristic function of $E_i$. Then we have $$\int_X s_n\,d\phi=\sum_i\alpha_i \phi(E_i) = \sum_i\alpha_i \int_{E_i}f\,d\mu=\sum_i\alpha_i \int_X\chi_{E_i}f\,d\mu=\int_X\sum_i\alpha_i\chi_{E_i}f\,d\mu,$$ where the last term is simply $\int_Xs_nf\,d\mu$. Since $\{s_n\}$ is an increasing sequence and $f$ is non-negative, clearly $\{s_nf\}$ is also increasing. Therefore, by MCT again, $$\lim_{n\to \infty} \int_X s_nf\,d\mu=\int_X\left(\lim_{n\to \infty} s_n f \right)\,d\mu=\int_Xgf\,d\mu,$$ which concludes the proof.
Is this correct? Can it be simplified or improved? Thanks a lot!